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Calculate the Energy that Can Be Obtained from 1 Kg of Water Through the Fusion Reaction 2h + 2h → 3h + P. Assume that 1.5 × 10−2% of Natural Water is Heavy Water D2o (By Number of Molecules) - Physics

Sum

Calculate the energy that can be obtained from 1 kg of water through the fusion reaction 2H + 2H → 3H + p. Assume that 1.5 × 10−2% of natural water is heavy water D2O (by number of molecules) and all the deuterium is used for fusion.

(Use Mass of proton mp = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)

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Solution

Given :-

18 g of water contains `6.023 xx 10^23` molecules.

`∴ "1000 g of water" = (6.023 xx 10^23 xx 1000)/18 = 3.346 xx 10^25` molecules

`% "of deuterium" = 3.346 xx 10^25 xx 0.015/100 = 0.05019 xx 10^23`

Energy of deuterium = `30.4486 xx 10^25`

`= [2 xx m(""^2H) - m(""^3H) - m_p]c^2`

`= (2 xx 2.014102  "u" - 3.016049  "u" - 1.007276   "u")c^2`

`= 0.004879 xx 931 "MeV"`

`= 4.542349" MeV"`

`= 7.262 xx 10^-13 "J"`

Total energy = `0.05019 xx 10^23 xx 7.262 xx 10^-13  "J"`

 = 3644 MJ

Concept: Nuclear Energy - Fission
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 24 The Nucleus
Q 53 | Page 444
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